问题及解答

\[
\int\frac{1+\cos x}{1+\sin^2 x}dx
\]

\[
\begin{split}
\int\frac{1+\cos x}{1+\sin^2 x}dx &=\int\frac{1}{1+\sin^2 x}dx+\int\frac{\cos x}{1+\sin^2 x}dx\\
&=\int\frac{\frac{1}{\cos^2 x}}{\frac{1}{\cos^2 x}+\tan^2 x}dx+\int\frac{d\sin x}{1+\sin^2 x}\\
&=\int\frac{\sec^2 x dx}{\sec^2 x+\tan^2 x}+\int\frac{d\sin x}{1+\sin^2 x}\\
&=\int\frac{d\tan x}{1+2\tan^2 x}+\int\frac{d\sin x}{1+\sin^2 x}\\
\end{split}
\]

前者令 $u=\tan x$, 后者令 $v=\sin x$.

\[
\begin{split}
\int\frac{d\tan x}{1+2\tan^2 x}&=\int\frac{du}{1+2u^2}=\frac{1}{\sqrt{2}}\int\frac{d(\sqrt{2}u)}{1+(\sqrt{2}u)^2}\\
&=\frac{1}{\sqrt{2}}\arctan(\sqrt{2}u)+C\\
&=\frac{1}{\sqrt{2}}\arctan(\sqrt{2}\tan x)+C
\end{split}
\]

\[
\begin{split}
\int\frac{d\sin x}{1+\sin^2 x}&=\int\frac{dv}{1+v^2}\\
&=\arctan v+C\\
&=\arctan(\sin x)+C
\end{split}
\]